document.write( "Question 969593: the perimeter of a rectangle is 150 yards. The width is 15 more than 1/3 the length. What is the area of the rectangle? \n" ); document.write( "

Algebra.Com's Answer #592419 by Boreal(15235) $\"\"$ $\"About$

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1350 sq yds. ANSWER\r
\n" ); document.write( "\n" ); document.write( "Perimeter =2L +2W or 2(L+W)\r
\n" ); document.write( "\n" ); document.write( "150=2(L+W) You don't need to divide by 2, but it makes the numbers smaller.
\n" ); document.write( "Divide by 2,
\n" ); document.write( "75=L+W\r
\n" ); document.write( "\n" ); document.write( "let length=x
\n" ); document.write( "width is (x/3)+15\r
\n" ); document.write( "\n" ); document.write( "x+(x/3)+15=75
\n" ); document.write( "Subtract 15 from both sides.
\n" ); document.write( "x+(x/3)=60\r
\n" ); document.write( "\n" ); document.write( "Multiply by 3 to clear the fractions.\r
\n" ); document.write( "\n" ); document.write( "3x +x =180
\n" ); document.write( "4x=180
\n" ); document.write( "Divide by 4\r
\n" ); document.write( "\n" ); document.write( "x=45
\n" ); document.write( "(x/3)=15
\n" ); document.write( "(x/3)+15=30\r
\n" ); document.write( "\n" ); document.write( "L=45 yards
\n" ); document.write( "W=30 yards
\n" ); document.write( "Perimeter =2(45)+2(30)=150
\n" ); document.write( "Area=L*W=45yds*30 yds=1350 sq yds.
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